# Thread: Continuous Branches of log z

1. ## Continuous Branches of log z

Hey everyone.

I am trying to define a continuous branch of log z in the
complex plane slit along the positive axis: C\{[0,i*infinity]}

I understand that we need to make a branch cut because log z will not be continuous on all of C. When the cut is the negative real axis I think I understand. And I also realize that the the cut can be any ray by my intuition. But I don't understand how to put it down as a function in this case.

2. Originally Posted by robeuler
Hey everyone.

I am trying to define a continuous branch of log z in the
complex plane slit along the positive axis: C\{[0,i*infinity]}

I understand that we need to make a branch cut because log z will not be continuous on all of C. When the cut is the negative real axis I think I understand. And I also realize that the the cut can be any ray by my intuition. But I don't understand how to put it down as a function in this case.
Define $\log z = \ln |z| + i \arg (z)$ where $\theta = \arg z$ is defined angle on $-\tfrac{3\pi}{2} < \theta \leq \tfrac{\pi}{2}$

3. Originally Posted by ThePerfectHacker
Define $\log z = \ln |z| + i \arg (z)$ where $\theta = \arg z$ is defined angle on $-\tfrac{3\pi}{2} < \theta \leq \tfrac{\pi}{2}$
I see, it is more straightforward than I thought. Thanks a lot.

I have to show that
|cosh z|^2 = cos^2(y)+sinh^2(x)

I have tried writing out cosh z in terms of e^x and e^y so that I can take the modulus the old fashioned way (x^2+y^2)^(1/2).
(from the e-definition of cosh z and using that e^z=e^xcos(y)+ie^xsin(y))

but the resulting mess is overwhelming and I can't seem to split up the real and imaginary parts neatly.

4. Originally Posted by robeuler
I see, it is more straightforward than I thought. Thanks a lot.

I have to show that
|cosh z|^2 = cos^2(y)+sinh^2(x)

I have tried writing out cosh z in terms of e^x and e^y so that I can take the modulus the old fashioned way (x^2+y^2)^(1/2).
(from the e-definition of cosh z and using that e^z=e^xcos(y)+ie^xsin(y))

but the resulting mess is overwhelming and I can't seem to split up the real and imaginary parts neatly.
If I did not mess up the identity it should be,
$\cosh (x+iy) = \cosh (x) \cosh (iy) + \sinh (x) \sinh (iy) = \cosh x \cos y + i\sinh x \sin y$

5. Originally Posted by ThePerfectHacker
If I did not mess up the identity it should be,
$\cosh (x+iy) = \cosh (x) \cosh (iy) + \sinh (x) \sinh (iy) = \cosh x \cos y + i\sinh x \sin y$
The identity is correct.
so |coshz|^2= cosh^2x * cos^2y + sinh^2x * sin^2y.

how do I get rid of the cosh and the sin terms?

Edit: Solved: substitute 1-cos^2y for sin^2y and it follows pretty easily.

6. Here is one more I hope you might get to:

If f(z) is an analytic branch of cosh^(-1)(z), find f'(z)

Can I simply take the derivative of
cosh^-1(z)= ln(x+(x-1)^(1/2)*(x+1)^(1/2))?

in the end this equals 1/((x-1)^(1/2)*(x+1)^(1/2))